\(X\left ( t\right ) \) is Stationary: If all its statistics do not change with shift of origin
\(X\left ( t\right ) \) is Wide Sense Stationary: If the mean is constant, and \(R_{x}\left ( t,t+\tau \right ) =R_{x}\left ( \tau \right ) \)
where autocorrelation \(R_{x}\left ( \tau \right ) \) is defined as \(E\left [ x\left ( t\right ) x^{\ast }\left ( t+\tau \right ) \right ] .\) Note, if \(X\left ( t\right ) \) is real, then \(R_{x}\left ( \tau \right ) \) is real and even
Note: \(R\left ( x\right ) \) must be WSS if it is ergodic\(.\)So ergodic process has constant mean.
4 How to determine Hilbert transform of a signal?
input \(x\left ( t\right ) \). Find \(\hat {x}\left ( t\right ) \) which is Hilbert transform of \(x\left ( t\right ) \) defined as \(\hat {x}\left ( t\right ) =x\left ( t\right ) \otimes \frac {1}{\pi t}\)
An easy way is to first find \(\hat {G}\left ( f\right ) \) which is the Fourier transform of \(\hat {x}\left ( t\right ) \) and then inverse it to find \(\hat {x}\left ( t\right ) \)
Suppose we have a message \(m\left ( t\right ) \) that is sampled. Assume we have \(n\) bits to use for encoding the sample
levels. Hence there are \(2^{n}\) levels of quantizations. We want to find the ration of the signal to the noise
power. Noise here is generated due to quantization (i.e. due to the rounding off values of \(m\left ( t\right ) \) during
sampling).
This is the algorithm:
Input: \(n\), the number of bits for encoding, \(m_{p}\) absolute maximum value of the message \(m\left ( t\right ) \), the
pdf \(f_{X}\left ( t\right ) \) of the message \(m\left ( t\right ) \) is \(m\left ( t\right ) \) is random message or \(m\left ( t\right ) \) function if it is deterministic (such as
\(\cos \left ( t\right ) \))
Find the quantization step size \(S=\frac {2m_{p}}{2^{2}}\)
Find \(P_{av}\) of the error is \(\frac {1}{12}S^{2}\) where \(S\) is the step size found in (1), hence \(P_{av}=\frac {1}{12}S^{2}=\frac {1}{12}\left ( \frac {2m_{p}}{2^{2}}\right ) ^{2}=\frac {m_{p}^{2}}{3\times 2^{2n}}\)
If \(m\left ( t\right ) \) is random, find \(p_{av}=E\left ( m\left ( t\right ) \right ) ={\int }m^{2}\left ( t\right ) f_{X}\left ( t\right ) dt\), this is called the second moment of the pdf
Hence find \(SNR\) for noise quantisation comes down to finding the power in the message
\(m\left ( t\right ) \).
Examples: For sinusoidal message \(m\left ( t\right ) \), \(SNR_{db}=6n+1.761\). For random \(m\left ( t\right ) \) with PDF which is uniform distributed \(SNR_{db}=6n\), for
random \(m\left ( t\right ) \) which is AWGN. Do this later
8 How to determine coding of a number from quantization?
Given an analog value say \(x\) and given a maximum absolute possible value to be \(m_{p}\), and given the
number of bits available for coding to be \(N\), the following are the algorithm to generate the
quantized version of \(x\), called \(\hat {x}\)
8.1 sign magnitude
Input: \(x,m_{p},N\)
output: \(\hat {x}\)
Let \(\Delta =\frac {m_{p}}{2^{N-1}}\) called the step size
Let \(q=round\left [ \frac {abs\left ( x\right ) }{\Delta }\right ] \) which is the quantization level
If \(q\geq 2^{N-1}-1\) then \(q=2^{N-1}\) end if
if \(x<0\) then \(code=q+2^{N-1}\) else \(code=q\) endif
return \(code\) in base 2
8.2 ones complement
Input: \(x,m_{p},N\)
output: \(\hat {x}\)
Let \(\Delta =\frac {m_{p}}{2^{N-1}}\) called the step size
Let \(q=round\left [ \frac {abs\left ( x\right ) }{\Delta }\right ] \) which is the quantization level
If \(q\geq 2^{N-1}-1\) then \(q=2^{N-1}-1\) end if
If \(x>0\) then \(code=q\) else \(code=\left ( 2^{N}-1\right ) -q\) endif
return \(code\) in base 2
8.3 offset binary
Input: \(x,m_{p},N\)
output: \(\hat {x}\)
Let \(\Delta =\frac {m_{p}}{2^{N-1}}\) called the step size
Let \(q=round\left [ \frac {abs\left ( x\right ) }{\Delta }\right ] \) which is the quantization level
If \(x\geq -\frac {\Delta }{2}\) then
if \(q\geq 2^{N-1}-1\) then
\(\ \ \ \ \ q=2^{N-1}-1\)
end if
\(code=2^{N-1}+q\)
else
if \(q\geq 2^{N-1}-1\) then
\(\ \ \ \ \ \ q=2^{N-1}\)
end if
\(code=2^{N-1}-q\)
end if
return \(code\) in base 2
8.4 2’s complement
Input: \(x,m_{p},N\)
output: \(\hat {x}\)
Let \(\Delta =\frac {m_{p}}{2^{N-1}}\) called the step size
Let \(q=round\left [ \frac {abs\left ( x\right ) }{\Delta }\right ] \) which is the quantization level
If \(x\geq -\frac {\Delta }{2}\) then
if \(q\geq 2^{N-1}-1\) then
\(\ \ \ \ \ q=2^{N-1}-1\)
end if
\(code=q\)
else
if \(q\geq 2^{N-1}-1\) then
\(\ \ \ \ \ \ q=2^{N-1}\)
end if
\(code=2^{N}-q\)
end if
return \(code\) in base 2
9 How to derive the Phase and Frequency modulation signals?
Where \(\tilde {x}\left ( t\right ) \) is the complex envelope of \(x\left ( t\right ) \). For PM and FM,
the baseband modulated signal, \(\tilde {x}\left ( t\right ) \) has the form \(A_{c}e^{j\theta \left ( t\right ) }\) Hence the above becomes
The above is the general form for PM and FM. Now, for PM, \(\theta \left ( t\right ) =k_{p}m\left ( t\right ) \) and
for FM, \(\theta \left ( t\right ) =k_{f}\int _{0}^{t}m\left ( t_{1}\right ) dt_{1}\). Hence, substituting in (1) we obtain
11 How to quickly determine SNR\(_{i}\) from \(SNR_{c}\)?
First find \(SNR_{c}\), for to find \(SNR_{i}\) use the following
\(SNR_{i}=SNR_{c}\frac {B}{B_{T}}\), where \(B_{T}\) is the transmission bandwidth, and \(B\) is the baseband bandwidth. For \(AM\), \(B_{T}=2B\). For \(DSB-SC\), \(B_{T}=2B\). For \(DSB-SS\),
\(B_{T}=B.\)
12 How to determine figure of merit for DSB-SC using coherent detector?
Figure of merit, \(\gamma \) is defined as \(\frac {SNR_{o}}{SNR_{c}}\) where \(SNR_{o}\) is the signal-to-noise ratio on output from modulator, and \(SNR_{c}\) is
signal-to-noise ratio for the channel, assuming channel has AWGN added. The following diagram
shows the calculations. I used a coherent demodulator.
Question: Verify the above.
13 How to determine figure of merit for AM transmission using coherent detector?
14 How to determine figure of merit for AM using envelope detector?
We notice,
that for Large \(SNR_{i}\), this detector gives the same result as coherent detector.
For small \(SNR_{i}\), it is better to use the coherent detector than the envelope detector.
15 How to determine figure of merit for SSB using coherent detector?
The difference here is that SSB signal has transmission bandwidth \(B_{T}=B\) and not \(2B\) as in all the previous
signals. Assume we are working with upper sideband. Analysis is the same for lower sideband.
Where \(k\) is a constant. Usually \(\frac {A_{c}}{2}\) but we will leave it as \(k\) for now. \(\hat {m}\left ( t\right ) \) is the Hilbert transform of
\(m\left ( t\right ) \)